11. Hyperbolic groups and quasiconvex subgroups

Proof of Theorem 6: Let be a quasi-geodesic in a -hyperbolic space . We can replace by as in Lemma 8. Let , , and such that is maximal. We need to bound . Let be such that , and such that . Let such that , and such that . Finally, let be a path in obtained by concatenating , the section of the image of with endpoints and , and .

By construction, does not intersect , and part 3 of Lemma 8 shows that

.

Lemma 7 gives a bound on the length of paths based on the distance between their endpoints:

.

The left hand side of this inequality increases linearly in , while the right increases logarithmically, so that must be bounded above for the equality to hold, and clearly this upper bounded depends only on the constants , and . QED.

It now makes sense to make the following definition.

Definition: A finitely generated group is called (word-) hyperbolic if some (any) Cayley graph for is Gromov-hyperbolic. Equivalently, a group is hyperbolic if it acts properly discontinuously and cocompactly by isometries on a proper Gromov-hyperbolic metric space.

Examples:

a) Free groups
b) is not hyperbolic for .
c) Let be any closed hyperbolic manifold. Then is word-hyperbolic.
d) More generally, is word-hyperbolic for any with negative sectional curvature bounded away from .

Without getting into too much detail, we briefly mention the following theorem of Gromov as an indication of how general the class of hyperbolic groups really is.

Definition: A subspace of a geodesic metric space is quasiconvex if there exists a such that, for all and for all , .

Example: Consider with the -metric. Then the diagonal subgroup is not quasiconvex (though it is quasi-embedded).

Theorem 6 implies that this kind of poor behavior does not occur in hyperbolic space.

Corollary: Suppose that is a word-hyperbolic group and is a subgroup. Then is quasiconvex in some (any) Cayley graph of if and only if is finitely generated and is a quasi-embedding.

Proof: is immediate from Theorem 6. For the other direction, fix a generating set for , assume is quasiconvex in the Cayley graph of with constant , and let be in . Consider a geodesic in the Cayley path of from to , which we can take to be of the form for in . Let be the vertices of this geodesic, so .

By quasiconvexity, for each there exists in such that . Take and . Let , so . Let . Note that . For each , we have that and so . Therefore, is generated by , a finite set. Furthermore, we have shown that .

But it’s clear that (as each element of has -length at most ) so the inclusion of into is a quasi-isometric embedding. QED.

In light of this, the following definition makes sense

Definition: A subgroup of a hyperbolic group is called quasiconvex if it is a quasiconvex space of some (any) Cayley graph of .

Exercise 14: If is a retraction and is finitely generated, then the inclusion is a quasi-isometric embedding. (Hint: you can choose a generating set for such that or for all .

Example: Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a retract of a finite-index subgroup, so every finitely generated subgroup of a free group is quasiconvex.

3 comments

It seems me that the proof for theorem 6 is not complete: it only proves that the geodesic is in a D-neigboorhood of the quasi geodesic, but doesn’t prove the other inclusion, so that the the hausdorff distance is not bounded.
Am i clear enough?
thank you

Benjamin, thanks for the comment, and sorry for being so slow to reply. I think you’re right. I haven’t had a chance to think hard about it, but I don’t think there’s any difficulty with the other direction – a similar argument should work.